The angle between the circles $x^{2} + y^{2} - 4 x - 6 y - 3 = 0, x^{2} + y^{2} + 8 x - 4 y + 11 = 0$

\frac{π}{2}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{6}

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Solution

The correct option is D $\frac{π}{3}$
$x^{2} + y^{2} - 4 x - 6 y - 3 = 0$
$r_{1} = \sqrt{2^{2} + 3^{2} + 3} = \sqrt{16}$ and $c_{1} = (2, 3)$
$= 4$
and $x^{2} + y^{2} + 8 x - 4 y + 11 = 0$
$r_{2} = \sqrt{4^{2} + 2^{2} - 11} = 3$ and $c_{2} = (- 4, 2)$
$∴ cos (180 - θ) = \frac{r {_{1}}^{2} + r {_{2}}^{2} - (c_{1} c_{2})^{2}}{2 r_{1} r_{2}}$
$cos (180 - θ) = \frac{25 - 37}{2 \times 4 \times 3}$
$cos θ = \frac{1}{2}$
$θ = 60^{0}$
$\Rightarrow θ = \frac{π}{3},$ angle $b e t^{n}$ circles.

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